Phenols - Reimer-tiemann Reaction
Treating phenol with chloroform and aqueous sodium hydroxide followed by acidification introduces a new functional group on the ring, giving which major organic product?
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Solution
Salicylaldehyde
The Reimer-Tiemann reaction converts phenol into an ortho-hydroxy aromatic aldehyde using chloroform in the presence of alkali. Sodium hydroxide first deprotonates chloroform to form the highly reactive electrophile dichlorocarbene, which then attacks the electron-rich, activated ortho position of the phenoxide ring. The resulting benzal chloride intermediate is hydrolysed by the alkali, and acidification finally gives salicylaldehyde, the 2-hydroxybenzaldehyde. Salicylic acid is wrong because forming a carboxyl group requires the Kolbe-Schmitt reaction with carbon dioxide, not chloroform. Benzaldehyde cannot form because the phenolic hydroxyl is retained and the new group enters ortho to it, not replacing it. Anisole is an ether made by O-methylation and is unrelated to carbene chemistry, so it is eliminated. This named reaction is part of the NCERT phenol chemistry and frequently appears in JEE product-prediction questions. The ortho preference arises because the phenoxide oxygen can hydrogen bond with and stabilise the developing intermediate near the adjacent position, and dichlorocarbene is electrophilic enough to attack only the highly activated phenoxide rather than neutral phenol. A small amount of the para-hydroxy aldehyde forms as a minor product, but the ortho isomer salicylaldehyde dominates. As a plausibility check, dichlorocarbene delivers exactly one carbon bearing the eventual CHO group at the ortho site, consistent with salicylaldehyde as the product.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing oxygen
- Topic
- phenols - reimer-tiemann reaction
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Salicylaldehyde
The Reimer-Tiemann reaction converts phenol into an ortho-hydroxy aromatic aldehyde using chloroform in the presence of alkali. Sodium hydroxide first deprotonates chloroform to form the highly reactive electrophile dichlorocarbene, which then attacks the electron-rich, activated ortho position of the phenoxide ring. The resulting benzal chloride intermediate is hydrolysed by the alkali, and acidification finally gives salicylaldehyde, the 2-hydroxybenzaldehyde. Salicylic acid is wrong because forming a carboxyl group requires the Kolbe-Schmitt reaction with carbon dioxide, not chloroform. Benzaldehyde cannot form because the phenolic hydroxyl is retained and the new group enters ortho to it, not replacing it. Anisole is an ether made by O-methylation and is unrelated to carbene chemistry, so it is eliminated. This named reaction is part of the NCERT phenol chemistry and frequently appears in JEE product-prediction questions. The ortho preference arises because the phenoxide oxygen can hydrogen bond with and stabilise the developing intermediate near the adjacent position, and dichlorocarbene is electrophilic enough to attack only the highly activated phenoxide rather than neutral phenol. A small amount of the para-hydroxy aldehyde forms as a minor product, but the ortho isomer salicylaldehyde dominates. As a plausibility check, dichlorocarbene delivers exactly one carbon bearing the eventual CHO group at the ortho site, consistent with salicylaldehyde as the product.
This medium difficulty chemistry question is from the chapter organic compounds containing oxygen, covering the topic of phenols - reimer-tiemann reaction. It appeared in the 2025 exam.
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